Q29E - University Physics with Modern Physics

Question

The walls of a soap bubble have about the same index of refraction as that of plain water, n = 1.33. There is air both inside and outside the bubble. 

(a) What wavelength (in the air) of visible light is most strongly reflected from a point on a soap bubble where its wall is 290 nm thick? To what color does this correspond (see Fig. 32.4 and Table 32.1)? 

(b) Repeat part (a) for a wall thickness of 340 nm.

Step-by-Step Solution

Answer

1. Green, 514 nm
2. Orange, 603 nm

Step 1: Given

$$\begin{gathered} {\mathrm{n}}_{\text{soab bubbble }}={\mathrm{n}}_{\text{fllm }}=1.33,\mathrm{t}=290\mathrm{nm}, \mathrm{The} \mathrm{visible} \mathrm{light} \mathrm{wavelength}: \\ 400\mathrm{nm}<{\mathrm{\lambda }}_{\text{air }}<700\mathrm{nm},{\mathrm{t}}_2=340\mathrm{nm} \end{gathered}$$

Step 2: Ray diagram.

Step 3: Solving part (a) of the problem.

Since we need the most strongly reflected light from the bubble, so we need to find the most strongly constructive interference. 

The light falls on the bubble surface from the air, as you see below, and then the first ray reflects with a phase change (since the bubble has an index of refraction which is greater than that of the air). The red circle indicates a phase change. 

But when the light hits the second surface of the bubble, the second ray reflects without any phase change (since the air now is the boundary surface) 

Now we have two reflected rays with one phase change. This means that the thickness of the soap bubble (the film) that gives a constructive interference would be given by

                                            $2\mathrm{t}=\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}$                                                  (1)

We know , from Snell's law,

                                          $n_1{\lambda }_1=n_2{\lambda }_2$                 

 that Hence,

                      ${\mathrm{n}}_{\text{film }\mathrm{m}}{\mathrm{\lambda }}_{\text{film }}={\mathrm{n}}_{\text{air }}{\mathrm{\lambda }}_{\text{air }}$                               

And the wavelength inside the film is then given by

                     ${\lambda }_{fibm}=\frac{{\lambda }_{a\dot{r}}}{n_{film}}$                                        

Whereas $n_{air}$ = 1.0

 Hence,

                                   ${\lambda }_{fibn}=\frac{{\lambda }_{a\dot{r}}}{n_{flm}}$                            

Substitute into (1) 

                              $2\mathrm{t}=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{air}}{n_{fibn}}$                               

solving for 

                       ${\lambda }_{\text{air }}=\frac{2t{n}_{fibn}}{\left(m+\frac{1}{2}\right)}$                                                                               (2)

Substitute the given

                                          $$\begin{gathered} {\lambda }_{a\dot{r}}=\frac{2\times 290nm\times 1.33}{\left(m+\frac{1}{2}\right)} \\ {\lambda }_{a\dot{r}}=\frac{771.4nm}{\left(m+\frac{1}{2}\right)} \end{gathered}$$                    

Now we will substitute m = 0,1,2, 3, .. 

When m=0;

                                         $\begin{array}{r}{\lambda }_{a\dot{v}}=\frac{771.4\mathrm{nm}}{\left(0+\frac{1}{2}\right)}\\ {\lambda }_{a\dot{v}}=1542.8\mathrm{nm}\end{array}$                    

which is not among the visible light wavelengths, as you see in the given above, it is greater than the longest wavelength of the visible light 

When m=1.0

                                                  $\begin{array}{c}{\lambda }_{a\dot{x}}=\frac{771.4\mathrm{nm}}{\left(1+\frac{1}{2}\right)}\\ {\lambda }_{air}=514.3\mathrm{nm}\end{array}$           

And this wavelength is visible, but we need to continue to see which wavelength of the light is the most bright 

Hence, when m = 2.0

                                                  $\begin{array}{r}{\lambda }_{a\dot{r}}=\frac{771.4\mathrm{nm}}{\left(2+\frac{1}{2}\right)}\\ {\lambda }_{a\dot{r}}=308.6\mathrm{nm}\end{array}$             

which is not among the Visible light wavelengths, as you see in the given above, it is shorter than the shortest wavelength of the visible light 

It is obvious that if we plugged m = 3.0, the resultant wavelength will be shorter than the last result, so we stop here 

Therefore, the most strongly visible light that is reflected from the soap bubble is a green light which has a wavelength of

                                                          $\lambda =514\mathrm{nm}$

Step 4: Solving part (b) of the problem.

We will use equation (2) above and substitute the new thickness for it

                                           ${\lambda }_{a\dot{r}}=\frac{2t_2{n}_{fibn}}{\left(m+\frac{1}{2}\right)}$          

Substitute the given

                                           $$\begin{gathered} {\lambda }_{a\dot{ir}}=\frac{2\times 340nm\times 1.33}{\left(m+\frac{1}{2}\right)} \\ {\lambda }_{a\dot{ir}}=\frac{904.4nm}{\left(m+\frac{1}{2}\right)} \end{gathered}$$        

Now we will Substitute m = 0,1,2, 3, ……

When m = 0;

                                                 $\begin{array}{r}{\lambda }_{a\dot{r}}=\frac{904.4\mathrm{nm}}{\left(0+\frac{1}{2}\right)}\\ {\lambda }_{a\dot{\overrightarrow{r}}}=1809\mathrm{nm}\end{array}$      

which is not among the visible light wavelengths, as you see in the given above, it is greater than the longest wavelength of the visible light 

When m = 1.0

                                            $\begin{array}{r}{\lambda }_{a\dot{r}}=\frac{904.4nm}{\left(1+\frac{1}{2}\right)}\\ {\lambda }_{a\dot{b}}=603\mathrm{nm}\end{array}$         

And this wavelength is visible, but we need to continue to see which wavelength of the light is the most bright Hence, 

when m= 2.0

                                               $\begin{array}{r}{\lambda }_{a{\dot{r}}^r}=\frac{904.4\mathrm{nm}}{\left(2+\frac{1}{2}\right)}\\ {\lambda }_{a\dot{r}}=362\mathrm{nm}\end{array}$      

which is not among the visible light wavelengths, as you see in the given above, it is shorter than the shortest wavelength of the visible light 

It is obvious that if we plugged m= 3.0, the resultant wavelength will be shorter than the last result, so we stop here. 

Therefore, the most strongly visible light that is reflected from the soap bubble is an orange light which has a wavelength of

                                       $\lambda =603\mathrm{nm}$